
\prob{009B}{三元分式方程组II}

求方程组
\[ \left\{ \begin{aligned}
  \frac{4x^2}{4x^2 + 1} &= y \\
  \frac{4y^2}{4y^2 + 1} &= z \\
  \frac{4z^2}{4z^2 + 1} &= x \\
\end{aligned} \right. \]
的实数根。
\problabels{yellow/代数, green/方程相关问题}

\textit{JYH提供的题目。}

\ans{$x = y = z = 0$或$x = y = z = 1/2$}

\subsection{倒数}

若$x, y, z$有一个为0，显然有$x = y = z = 0$，是方程组的一组解。若$x, y, z$均不为0，则三方程两边取倒数得
\[ \frac1{4x^2} + 1 = \frac1y, \frac1{4y^2} + 1 = \frac1z, \frac1{4z^2} + 1 = \frac1x \]
三式相加，得
\[ \frac1x + \frac1y + \frac1z = 3 + \frac14\left(\frac1{x^2} + \frac1{y^2} + \frac1{z^2}\right) \]
由此，注意到
\[ \frac1{x^2} + \frac1{y^2} + \frac1{z^2} - 4\left(\frac1x + \frac1y + \frac1z\right) + 12 = 0 \]
配方，得
\[ \left(\frac1x - 2\right)^2 + \left(\frac1y - 2\right)^2 + \left(\frac1z - 2\right)^2 = 0 \]
因此$x = y = z = 1/2$。

综上，$x = y = z = 0$或$x = y = z = 1/2$。
